As far as I understand, the reason that GHC does not construct such proofs is that it can't express them in its internal proof language (System FC).
Iavor is quite right
It seems to me that it should be fairly straight-forward to extend FC to support this sort of proof, but I have not been able to convince folks that this is the case. I could elaborate, if there's interest.
Iavor: I don’t think it’s straightforward, but I’m willing to be educated. By all means start a wiki page to explain how, if you think it is.
I do agree that injective type families would be a good thing, and would deal with the main reason that fundeps are sometimes better than type families. A good start would be to begin a wiki page to flesh out the design issues, perhaps linked from http://hackage.haskell.org/trac/ghc/wiki/TypeFunctions
The main issues are, I think:
· How to express functional dependencies like “fixing the result type and the first argument will fix the second argument”
· How to express that idea in the proof language
Simon
From: glasgow-haskell-bugs-bounces at haskell.org [mailto:glasgow-haskell-bugs-bounces at haskell.org] On Behalf Of Iavor Diatchki
Sent: 26 December 2012 02:48
To: Conal Elliott
Cc: glasgow-haskell-bugs at haskell.org; GHC Users Mailing List
Subject: Re: Fundeps and type equality
Hello Conal,
GHC implementation of functional dependencies is incomplete: it will use functional dependencies during type inference (i.e., to determine the values of free type variables), but it will not use them in proofs, which is what is needed in examples like the one you posted. The reason some proving is needed is that the compiler needs to figure out that for each instantiation of the type `ta` and `tb` will be the same (which, of course, follows directly from the FD on the class).
As far as I understand, the reason that GHC does not construct such proofs is that it can't express them in its internal proof language (System FC). It seems to me that it should be fairly straight-forward to extend FC to support this sort of proof, but I have not been able to convince folks that this is the case. I could elaborate, if there's interest.
In the mean time, the way forward would probably be to express the dependency using type families, which I find tends to be more verbose but, at present, is better supported in GHC.
Cheers,
-Iavor
PS: cc-ing the GHC users' list, as there was some talk of closing the ghc-bugs list, but I am not sure if the transition happened yet.
On Tue, Dec 25, 2012 at 6:15 PM, Conal Elliott <conal at conal.net<mailto:conal at conal.net>> wrote:
I ran into a simple falure with functional dependencies (in GHC 7.4.1):
> class Foo a ta | a -> ta
>> foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool
> foo = (==)
I expected that the `a -> ta` functional dependency would suffice to prove that `ta ~ tb`, but
Pixie/Bug1.hs:9:7:
Could not deduce (ta ~ tb)
from the context (Foo a ta, Foo a tb, Eq ta)
bound by the type signature for
foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool
at Pixie/Bug1.hs:9:1-10
`ta' is a rigid type variable bound by
the type signature for
foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool
at Pixie/Bug1.hs:9:1
`tb' is a rigid type variable bound by
the type signature for
foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool
at Pixie/Bug1.hs:9:1
Expected type: ta -> tb -> Bool
Actual type: ta -> ta -> Bool
In the expression: (==)
In an equation for `foo': foo = (==)
Failed, modules loaded: none.
Any insights about what's going wrong here?
-- Conal
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